In Sudoku terms, a pair is two cells in the same house (row, column, or box) that have the same two candidates. No other cell in the same house can have either of these candidates.

In other words: If any two cells that are peers have the same two candidates (and only two candidates), those candidates can be eliminated in their shared peers.

Sometimes we find that more than two cells in a puzzle have the same two candidates. Moreover these cells can sometimes be linked in a chain to form a remote pair. The cells in the remote pair are not peers, but their candidate values can be eliminated in the cells that are their shared peers.

Note that the cells marked A and B each have the same two candidates, and are in the same row. A and B are a pair, as are B and C, and C and D. They form a chain:
A=B=C=D or
XY=XY=XY=XY
with two possible solutions:
4 5 4 5 or
5 4 5 4.

A and D are a remote pair. They are not peers, but in the solution one of them is <4>, and the other must be <5>. We can eliminate the candidates in their shared peers: <45> in R3C7, <4> in R4C8, and <5> in R6C8. This solves the puzzle.

Note that, in order to be a remote pair, the chain must have an even number of cells, and an odd number of links. If we have

XY=XY or XY=XY=XY=XY or XY=XY=XY=XY=XY=XY

we can say that one end is X and the other is Y, and we can eliminate X and Y from the shared peers of the pincer cells at the ends of the chain.

If, on the other hand, the chain has an odd number of cells, and an even number of links

XY=XY=XY or XY=XY=XY=XY=XY

we can only say that both ends of the chain are the same value. Both are X, or both are Y. Andrew Stuart (link below) calls this a complementary pair.

Above is the usual explanation of remote pairs. They are not very common, but when they do occur they are very easy to find.

But, there is much more! The rest of this thread explores the question: What can I do if two cells that are not peers have the same two candidates?

Credits

Remote pairs were apparently discovered by Andrew Stuart. His description is here:

House. Row, column, or 3x3 box.
Line. Row or column.
Peers. Cells that a single cell can "see". A single cell has 20 peers: Eight in its row, eight in its column, and another four in its box (that are not in its row or column).
Shared peers. Peers that two cells have in common. Two cells have at least two, and as many as 13 shared peers.[/i]

Last edited by keith on Sat Dec 29, 2007 3:22 pm; edited 3 times in total

A=B=C=D=E=F is a chain of strong links in <5>. In the solution, one of A or F is <5>. The other must be <4>, since A and F have the same two candidates <45>.

A number of people seem to know this, but I have not been able to find it written down anywhere: If two cells that are not peers have the same two candidates, and can be connected by a chain of an odd number of strong links in one of the candidates, the two cells are a remote pair.

A=B=C=D is a chain of strong links on <2>, connecting two cells with the same pair of candidates, <12>. We can eliminate <12> from all their shared peers, including <1> in R8C8. A and D are a remote pair.

[Yes, I understand that, in this example, A and D can also be connected by a chain of strong links in <1>. That does not affect the logic described above.]

Last edited by keith on Sat Dec 29, 2007 9:36 pm; edited 1 time in total

We now continue the discussion: If two cells that are not peers each have the same two candidates, how might this be useful?

If the cells are a remote pair, they must be connected by an odd number of strong links in one candidate. If the cells are connected by an even number of strong links, they are a complementary pair. It is It is often said that a complementary pair is not useful. This is not true!

A and C are a complementary pair, joined via their strong links on <3> to B. C is extended to D by a strong link on <6>. We can eliminate <6> in R1C3.

So: If two cells that are not peers have the same two candidates and can be connected by an even number of strong links in one candidate, they are a complementary pair. Those two cells will have the same value in the solution. If either end can be extended by a strong link on the other candidate, that candidate can be eliminated in the shared peers of the ends of the extended chain.

To summarize, the M-wing is

XY=X=XY=Y

and we can eliminate Y from the shared peers of the end cells.

I have called this an M-Wing, for it is the simplest case of Medusa coloring. Also, it is similar to a W-Wing, which is the next topic of discussion.

More than two cells in the same house have the same candidate value. Any two of these cells are weakly linked. In the solution, at least one of them is not true.

B. Strong link

Two (and no other) cells in the same house have the same candidate value. These cells are strongly linked. In the solution, one of them has the candidate value, the other does not.

C. Pair

Two cells in the same house have the same two (and only two) candidate values. These cells are strongly linked in both candidates. In the solution, one cell is the first candidate, the other is the second.

D. Classic Remote Pair

Two cells that are not peers have the same two candidate values. They can be linked by a chain of cells that:

1. All have the same pair of candidates.

2. Has an even number of cells, which is equivalent to an odd number of links.

The ends of the chain

XY=XY=XY=XY

are a remote pair. The chain is comprised of four cells and three strong links on both X and Y.

E. General Remote Pair

Two cells that are not peers have the same two candidate values. They can be linked by a chain of cells that:

1. Comprises strong links on one of the candidates.

2. Has an even number of cells, which is equivalent to an odd number of links.

The ends of the chain

XY=X=X=XY

are a remote pair. The chain is comprised of four cells and three strong links on X.

F. Classic Complementary Pair

Two cells that are not peers have the same two candidate values. They can be linked by a chain of cells that:

1. All have the same pair of candidates.

2. Has an odd number of cells, which is equivalent to an even number of links.

The ends of the chain

XY=XY=XY

are a complementary pair. The chain is comprised of three cells and two strong links on both X and Y.

G. General Complementary Pair

Two cells that are not peers have the same two candidate values. They can be linked by a chain of cells that:

1. Comprises strong links on one of the candidates.

2. Has an odd number of cells, which is equivalent to an even number of links.

The ends of the chain

XY=X=XY

are a complementary pair. The chain is comprised of three cells and two strong links on X.

A and C have the same two candidates. They are not a remote pair, they are not a complementary pair. All you can say is: If A is <4>, C is <4>. (There is a weak link A-B, and a strong link, B=C, on <4>.)

But, there is also a strong link on <3> in C=D.

The logic for the elimination is:

Either:

1. A is <3>.

Or:

2. A is <4>, C is <4>, D is <3>.

Any cell that sees both A and D cannot be <3>. In particular, we can take out <3> from R5C6.

This is from the March 16 VH. Note the potential W-Wing on 25 in boxes 2 and 9. There is no connecting strong link that I can see. But the 5 in r5c7 sees the pair in box 9; by coloring this with two strong links, you get to r6c4, which sees the pair in box 2. I thought this was the equivalent of a strong link connection on 5, which would makes the pincers = 2 on the two pairs. But then testing the pairs shows that they are both 2 or both 5, thus no pincer effect.

Where have I gone astray?

(The grid is apparently not in error, since I can use the 89-58-59 XY-Wing to take out the 2 from r3c4 and successfully complete the puzzle).

The connecting chain must be one, or three, or five, ... links. Yours, bcd, is only two. This is not a W-wing.

But, it is a potential M-wing. Note the coloring chain on <5> abcde. Since a and e have the same two candidates, they are a complementary pair: In the solution, they have the same value. You can complete the M-wing by adding a strong link on 2: af (or ag). The pincers in ef or eg are on <2>; you can take out <2> in R8C5.

The idea of the M-wing is this: Suppose you have a complementary pair, XY ... XY. They are connected in some way so that you know they have the same value in the solved puzzle. If you can append a strong link in either X or Y to either cell, you have a chain with pincers X or Y.

You may then have this chain:

XY ... (X) ... XY = aY

where a is anything and = is a strong link in Y. The ... (X) ... simply means that there is some logic concerning X that leads you to conclude the XY cells have the same value.

Nataraj and / or Asellus pointed out that the chain can be

XY ... (X) ... bXY = aY

where b is anything. The logic is a little different, but the conclusion still is: One, or both, of the pincers is Y.

A and C have the same two candidates. They are not a remote pair, they are not a complementary pair. All you can say is: If A is <4>, C is <4>. (There is a weak link A-B, and a strong link, B=C, on <4>.)

But, there is also a strong link on <3> in C=D.

The logic for the elimination is:

Either:

1. A is <3>.

Or:

2. A is <4>, C is <4>, D is <3>.

Any cell that sees both A and D cannot be <3>. In particular, we can take out <3> from R5C6.

Hi Keith, sorry to join the party so late. I just found out what an M-wing is. It seems to me that (at least in this particular case) your half M-wing creates a continuous loop:
(3)r6c7 = r6c6 - (3=4)r4c5 - r4c7 = (4-3)r6c7
This not only implies that r5c6<>3, but that r4c4<>4. Is this already well-known?
By the way, the same things holds for ravel's original allowing you to also eliminate 8 from r7c157.

When the ends (pincers) of an m-wing ("classic","half" or "generalized", it doesn't matter) share a house ("see" each other), the loop is completed and all weak links turn into strong links, allowing additional eliminations.

Definitely not something that was "well known" to me, and I'll be on the lookout for those situations from now on!

When the ends (pincers) of an m-wing ("classic","half" or "generalized", it doesn't matter) share a house ("see" each other), the loop is completed and all weak links turn into strong links, allowing additional eliminations.

Definitely not something that was "well known" to me, and I'll be on the lookout for those situations from now on!

Hey nataraj,

It's nice to be back. I've missed you guys and I'm glad the names I see around here are the same as before.

Your rephrasing is exactly what I had in mind and I appreciate your more precise statement. I should note that this idea is a direct result of http://www.sudoku.com/boards/viewtopic.php?p=64249#64249thread(edit: Link fixed 3/29/2011 - keith.)
over on the player's forum started by Bud. In that thread he presents what he calls a double w-wing. In the end, it is simply cells {a b}, {a b} that are connected by w-wings on both digits. My observation (which I again assume is well-known) is that this gives you a continuous loop (so all weak links turn into strong links) regardless of the relative positions of the two cells (i.e., they don't have to see each other).

When the ends (pincers) of an m-wing ("classic","half" or "generalized", it doesn't matter) share a house ("see" each other), the loop is completed and all weak links turn into strong links, allowing additional eliminations.

Definitely not something that was "well known" to me, and I'll be on the lookout for those situations from now on!

Hey nataraj,

It's nice to be back. I've missed you guys and I'm glad the names I see around here are the same as before.

Your rephrasing is exactly what I had in mind and I appreciate your more precise statement. I should note that this idea is a direct result of thread over on the player's forum started by Bud. In that thread he presents what he calls a double w-wing. In the end, it is simply cells {a b}, {a b} that are connected by w-wings on both digits. My observation (which I again assume is well-known) is that this gives you a continuous loop (so all weak links turn into strong links) regardless of the relative positions of the two cells (i.e., they don't have to see each other).

I've been staring at this, reaching towards a statement something like that made by nataraj. I was sort of leaning towards some kind of "double M-wing reversible loop" explanation. (I have yet to read Bud's thread.)

Quote:

cells {a b}, {a b} that are connected by w-wings on both digits

In that case, the cells {a b} are a remote pair. Obviously. Why this is remarkable, I do not know. The first W-wing destroys the first candidate, the second the second. Knowing both of them together does not add anything to knowing them separately.

Let me read, and think, some more. When there is a post that includes ravel, re'born, and nataraj in the same breath, I know I am way over my head!